Quote:
Originally Posted by mart_r
For example, for large enough odd sigma(n)n, it's quite easy to find a possible n. It doesn't take long to find a prime p such that q=(sigma(n)n1p) is also prime and n=p*q is found.
The problem now is determining whether this is the smallest solution (mostly it isn't), and how efficiently can it be found?

Hello,
I work on this problem and you can see my results on the link :
http://www.aliquotes.com/nombre_ante...aliquotes.html
But I'm sorry, it's in french !
Exactly, I want to find the number of aliquot antecedents of the integers (odd and even). My program is written in Mathematica langage and it is very fast. This program finds all the aliquot antecedents of the integers, but I only count them for each integer and not memorize them.
You can also dowload some files on my aliquot sequences data base :
http://www.aliquotes.com/aliquote_base.htm
For information, since february 23th 2013, I'm computing a big program to find the number of aliquot antecedents of the integers from 1 to 50 000 000. I think that the end of this very long calcul will be in january 2014. And at this date, I will put the result on the same internet link.
I hope I help you !
JeanLuc Garambois